Let $X_0, X_1, X_2, ..., X_n$ each be non-identical independent random variables.

Let $x_0, x_1, ... , x_n$ be possible values of each of those random variables.

Let $\operatorname{Pdf}_{x0}(x_{0}), \operatorname{Pdf}_{x1}(x_{1}), ... , \operatorname{Pdf}_{x_n}(x_{n})$ represent each of random variables corresponding probability density functions.

$X_0$ has probability density function $\operatorname{Pdf}_{x0}(x_0)$ ,

$X_1$ has probability density function $\operatorname{Pdf}_{x1}(x_1)$ ... etc...

Lets say we know explicitly, analytically, exactly what all the $Pdf$'s are above.

Now Let $Y = \max( X_0, X_1, X_2, ..., X_n )$ which has probability density function $Pdf_y(y)$.

What is $\operatorname{Pdf}_y(y)$ ?

How do I calculate it with the information above?

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    $\begingroup$ Check out en.wikipedia.org/wiki/Order_statistic. $\endgroup$ – Andrew Jun 9 '16 at 22:23
  • $\begingroup$ They are not identical random variables. $\endgroup$ – D Adams Jun 9 '16 at 22:24
  • $\begingroup$ @probablyme clarified $\endgroup$ – D Adams Jun 9 '16 at 22:27
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    $\begingroup$ The computation is similar as for the case where the random variables are identical. The only difference is that you can not take the $n$th power of the distribution function. Instead that you have to consider the product of all the distribution functions seperately. $\endgroup$ – Cavents Jun 9 '16 at 22:31

The CDF of a maximum is easy (for independent things):

$$P\left(\max(X_0,X_1, \ldots,X_n) \leq c\right) = P(X_0,X_1,\ldots,X_n \leq c) = P(X_0 \leq c) \times P(X_0 \leq c) \times \ldots \times P(X_n \leq c) .$$

That is, the CDF at $c$ of the maximum of independent variables is the product of the CDF's of the variables at $c$.

To get the PDF, simply differentiate the CDF with respect to $c$.

  • $\begingroup$ The joker doesn't stand a chance! $\endgroup$ – D Adams Jun 9 '16 at 23:11

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